Question

A 12 kg rock is dropped off a 2 story house with a flat roof( 1 story=3 m). If the stone lands on a large spring, compressing the spring by 0.5 m before coming to rest, what is the spring constant k of the spring? The spring stands vertically on level ground and has an equilibrium length of 1m. Show all work please.

Answer 1

first, need to find out the stone speed just before it touches the spring

v^2 = u^2 +2*g*h ,u = 0 , So v= sqrt(2*g*h)

here h = 2*3- 1 = 5 m so v = sqrt(2*9.8*5) = 9.89949494 m/s

now apply energy conservation here  

KE+ PE initial = KE+ PE final

take PE refrence level at ground  

here h= 1 = lengh of spring

spring start to compress upto x so rock height after it stopped = (h-x) = 1-0.5 =0.5m

1/2*m*v^2 +m*g*h = 0 + m*g( h-x) + 1/2*k*x^2

0.5*12*9.899^2 +12*9.8*1 = 12*9.8*(1-0.5) +0.59*k*0.5^2

from here  

k≈4384.69  N/m answer 

let me know in a comment if there is any problem or doubts